NTA Abhyas JEE Main2020ChemistrySolutionsPractice
A 0.10 M solution of a monoprotic acid (d = 1.01 g/cm 3 ) is 5% ionized. What is the freezing point of the solution ? The molecular weight of the acid is 300 g/mol and K f (H 2 O) = 1.86 o C/m.
Options
- A-0.189 o C
- B-0.194 o C
- C-0.199 o C
- DNone of these
Correct answer
C. -0.199 o C
Step-by-step solution
Mass of 1 litre of solution = 1010 g Mass of solvent = 1 0 1 0 - 3 0 0 × 0.1 ⇒ 9 8 0 g Molality(m) = 0.1 0.98 ⇒ 0.102 ; Δ T f = K f . m . i = 1 + α K f · m Δ T f = 1.05 × 1.86 × 0.102 = 0.199 o C ; T f = 0 - 0.199 = - 0.199 o C